In order to find the sum of two polynomials, we must first set up the operation and identify the like terms.

The like terms in these polynomials are the squared variable, the single variable, and the constant terms.

The sum of these polynomials is equal to the following expression:

In order to find the difference of two polynomials, first identify like terms. The like terms in these polynomials are the squared variable, the single variable, and the constant terms. 

Remember, distribute the negative sign to all terms within the parentheses.

Solve.

The correct answer is .

To find the product of two polynomials, first set up the operation.

Now, multiply each term in the first polynomial by each term in the second polynomial. One way to remember to work through this type of multiplication is by using the acronym FOIL: First, Outside, Inside, Last. You multiply the first numbers in each parenthetical with one another, then the numbers on the outside of the entire list (the first number in the first parenthetical and the last one in the second parenthetical), then the inside numbers (the two middle ones, the last number in the first parenthetical and the first number in the second parenthetical), and finally the last numbers in each parenthetical.

When exponential terms with the same bases are multiplied together, you add the exponents. 

Combine like terms to arrive at the solution.

When the two polynomials are multiplied together, they equal .

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (6, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (5, or b in the standard quadratic formula). Because their product is positive (6) and the sum is positive, that must mean that they both have positive signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 2 and 3, as the product of 2 and 3 is 6, and sum of 2 and 3 is 5. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-9, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-9) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 3 and -3, as the product of 3 and -3 is -9, and sum of 3 and -3 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (2, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-3, or b in the standard quadratic formula). Because their product is positive (2) and the sum is negative, that must mean that they both have negative signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 2, as the product of 1 and 2 is 2, and sum of 1 and 2 is 3. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (2, or b in the standard quadratic formula). Because their product is positive (1) and the sum is positive, that must mean that they both have positive signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 1, as the product of 1 and 1 is 1, and sum of 1 and 1 is 2. So, this results in the expression's factored form looking like...

From here, set the binomial equal to zero and solve for .

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zero of the function is,

To find the zeros of the function using factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (4, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-4, or b in the standard quadratic formula). Because their product is positive (4) and the sum is negative, that must mean that they both have negative signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are -2 and -2, as the product of -2 and -2 is 4, and sum of -2 and -2 is -4. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for . Since the binomials are the same, there will only be one zero.

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zero of the function is,

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-1) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -1, as the product of 1 and -1 is -1, and sum of 1 and -1 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (3, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (4, or b in the standard quadratic formula). Because their product is positive (3) and the sum is positive, that must mean that they both have positive signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 3, as the product of 1 and 3 is 3, and sum of 1 and 3 is 4. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore, the zeros are,